Digital Circuit Simulation Questions, Exercises of Digital Logic Design and Programming

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DIGITAL

ASSIGNMENT 1

CSE 1003 – DIGITAL LOGIC AND DESIGN

SUBMITTED BY: ROHITH PILLAI

REGISTRATION NUMBER:19BCE

SLOT: L29 + L

LAB: SJT 122

1. Aim:

To get familiarized with MultiSim software. Write the procedure for designing a

circuit.

Procedure:

1. Open MultiSim and open a new page.

2. To select components to place, go to Place->Component.

3. Select the appropriate category of the component under Group.

4. Select the required component and click OK.

5. Place the component by left clicking wherever desired.

6. Rotate components by right clicking on them, if required.

7. Connect all the components with wire by dragging the mouse from one

component to the other.

8. Simulate the circuit by clicking the green play button.

9. End the simulation by clicking on the red stop button.

To implement all logic gates using NAND/ NOR gates and verify their truth table.

Boolean expression:

USING NAND

OR – ((A∙BA)’∙B(B∙BB)’)’

AND – ((A∙BB)’∙B(A∙BB)’)’

NOT – (A∙BA)’

NOR – (((A∙BA)’∙B(B∙BB)’)’ ∙B( (A∙BA)’ ∙B(B ∙BB)’)’)’

XOR – ((A ∙B(A ∙BB)’)’ ∙B(B ∙B(A ∙BB)’)’)’

XNOR – ((A ∙BB)’ ∙B((A ∙BA)’ ∙B(B ∙BB)’)’)’

USING NOR

OR – ((A+B)’+(A+B)’)’

AND – ((A+A)’+(B+B)’)’

NOT – (A+A)’

XOR – ((A+B)’+((A+A)’+(B+B)’)’)’

Truth Table:

USING NAND

A B A

B

A

∙B

B

A

(A

+B

A ∙BB

’+A’

∙B B

A ∙BB

+A’

∙B B’

((A ∙BA)

’ ∙B(B ∙B

B)’)’

((A ∙BB)

’ ∙B(A ∙B

B)’)’

(A

∙B

A)’

(((A ∙BA)’ ∙B(B ∙BB)

’)’ ∙B((A ∙BA)’ ∙B(B

∙B B)’)’)’

((A ∙B(A ∙BB)’

)’ ∙B(B ∙B(A ∙B

B)’)’)’

((A ∙BB)’ ∙B((

A ∙BA)’ ∙B(B ∙B

B)’)’)’

USING NOR

A B A+

B

A∙B

B

A’ A∙BB’+A’

∙B B

((A+B)’+

(A+B)’)’

((A+A)’+

(B+B)’)’

(A+A)

((A+B)’+((A+A)’+

(B+B)’)’)’

Output:

USING NAND

4. Aim:

To verify De Morgan’s, Distributive, and Associative theorems.

Boolean expression:

De Morgan’s Theorems – (A∙BB)’=A’+B’

(A+B)’=A’ ∙ B’

Distributive Law – A+(B∙BC) = (A+B)∙B(A+C)

Associative Law – A+(B+C) = (A+B)+C

Truth Table:

DE MORGAN’S THEOREMS

A B A’ B’ A∙BB A+B (A∙BB)’ A’+B’ (A+B)’ A’∙BB’

DISTRIBUTIVE LAW

A B C B∙BC A+(B∙BC) A+B A+C (A+B)∙B(A+

C)

ASSOCIATIVE LAW

A B C B+C A+(B+C) A+B (A+B)+C

Output:

DE MORGAN’S THEOREMS

ASSOCIATIVE LAW

5. Aim:

To implement:

a. 3-bit Binary to Gray

b. 3-bit Gray to Binary

c. BCD to Excess-

Boolean expression:

BINARY TO GRAYCODE

G2=B

G1=B2 XOR B

G0 =B1 XOR B

GRAYCODE TO BINARY

B2 =G

B1 =G2 XOR G

B0 =G2 XOR G1 XOR G

B.C.D. TO EXCESS-

E3 =A+(B ∙BD)+(B ∙BC)

E2=(B ∙BC’ ∙BD’)+(B’ ∙BD)+(B’ ∙BC)

E1= (C’ ∙BD’)+(C ∙BD)

E0 =D’

Truth Table:

BINARY TO GRAYCODE

B2 B1 B0 G2 G1 G

GRAYCODE TO BINARY

G2 G1 G0 B2 B1 B

GRAYCODE TO BINARY

B.C.D. TO EXCESS-